in-exercises-17-34-sketch-the-graph-of-the-quadratic-function-without-using-a-graphing-utility-id-17

Question

In Exercises 17-34, sketch the graph of the quadratic function without using a graphing utility. Identify the vertex, axis of symmetry, and x-intercept(s).

$ f(x) = \frac{1}{4} x^2 - 2x - 12 $

J Hardin · Accepted Answer

start off here by writing this in the form. And the reason for doing that is that the Vertex will be h k. So here we have one over four ex clear minus a dicks. Parentheses minus twelve. Now we&#x27;LL go ahead and complete the square. So half of eight is four square that you get sixteen. But notice we didn&#x27;t just add in the forth the sixteen here because we have one fourth. So we really added it. Four. So we&#x27;ll make up for this vice attracting for And then we&#x27;re still completing the square And there we go We see that the vertex is the point four minus sixteen So for here in the eggs And then I&#x27;m going down by twos in the white direction And then next up we have access of symmetry This has always given by x equals age So in our case, that&#x27;s for So if I come over here to the graph, that would be more or less This photo Klein right here passing through the Vertex So that equation X equals four. And also we have X intercepts do this in blue. So as usual we take our function f so we can use this or we can use this expression over here. Set that equals zero. Push the sixteen to the other side. Multiply that four over. So square that plus or minus eight. So X equals four plus or minus eight. And go ahead and plug in both of those values. The plus in the minus. And that will give you it&#x27;s fallen minus four. So we know that our graph would pass through this point here. And by the way that I draft this out, we can&#x27;t see the X equals twelve. But that should also be planet on the graph. And then you would just go ahead and possibly plot in a few more points, for example, f of zero negative twelve. So we have some points here, so and that&#x27;s a graph.

J Hardin · Answer

Let&#x27;s graph this problem my first thirty with the graph of X Square and then we&#x27;LL multiply that by a half And then finally we&#x27;LL subtract by four. So we&#x27;re using transformations of the problem X Square. So let&#x27;s go ahead and give a rough sketch and red. So our usual Graff X squared. Let&#x27;s go ahead and put that in it with a dotted line. So this is not our final answer. Take a step back here in a little sloppy, some more or less a rough sketch of your proble X squared. Next step in blue. It&#x27;s the good and multiply that by a half. So this means you take each point on red, take the Y value and multiply it by a half. So over here we see the Y values for multiply That by a half, gives you two. Here we see one multiplying that by a half, gives you a half zero times a half a zero. One times a half is one half and once again, four times a half is to So this is the bluegrass one half X squared and out. Finally, we&#x27;re ready for the final graph f and so we just take the blue graph and we do a minus for So this means we take each choir value on the blue graft and shifted down for So this point right here is too and subtracting for from that will put you at minus two. No, this point here was zero. Subtracting four puts you at minus four. Here we have two again subtracting for and this point was a half subtract for from that Excuse me there. And that should give you negative seven over too. Let&#x27;s say three point five. And of course, we could go ahead and plug it. More points to get a more accurate looking graph. For example, if you plug in X equals three or minus three. So if you plug in three year minus three, you get a half. So let&#x27;s just go ahead and for more detail, which is plug in that last point. Okay, so here&#x27;s FX and Black. So that&#x27;s the first task. Sketch the raft. Now let&#x27;s identify the Vertex. Here we see the vortex is the point zero minus four and therefore the access of symmetry is always you just said x equal to the X value of the Vertex so that&#x27;LL just be the y axis. This is the access of symmetry and finally, the ex intercepts. So these we obtained by setting f of X equals zero Lisa room over here to do the algebra. So now you have one half X squared minus four equals zero So that gives you x square. So add the four over and then multiply by two to get X squared is eight and then finally X equals plus or minus run of a which is plus or minus too radical, too. So that will give us the ex intercepts Elice, potties and green And then over here, the minus one.

J Hardin · Answer

first write F in the Vertex form. So this will be something like and then our vertex will be the point. I&#x27;m scared. Okay, So let me go ahead and pull out a negative sign just from the first two terms. Now go ahead and take half of four and square that and you get four back. But notice we didn&#x27;t just ad for and really, we didn&#x27;t mind us for here, So make up for that by adding forthe variant. Go ahead and complete that square. And there we are. Theirs are for takes form so many ways to could go ahead and get the rest of the graph. You go. You can go ahead and just plugging a few more points here. Or you can just see that this is a transformation of X squared. But if you want to go ahead and just plugging a few points here, you could plug in negative one. You get four. He plugged in negative three. You also good for and then half of zero. That should give you one. And if you do f of minus four, that should also give you one. So there&#x27;s a few points and let&#x27;s go ahead and give a rough sketch here and less Label of Vertex. Negative to five. Okay, so photograph a rough sketch. Okay, so then completes this part. Sketched the graph now identified. The verse six where we have that access of cemetery. This has always given my X equals h. In our case, H is minus two. So we&#x27;LL say X equals minus two. This is the axis and finally, the ex intercepts. So we&#x27;re trying to find out what these two points are. So we&#x27;LL go ahead and set f equal to zero. So that becomes Take the square ruler on both sides. Don&#x27;t forget the plus and minus and then subtract it too. So these will be our X intercepts.

J Hardin · Answer

for this problem will sketch the wrath of the quadratic function. So perhaps the best way to do this is to start off with the parent function X Square. The basic problem. The first one that we&#x27;ve learned autograph and to use the sequence of transformations once that by the time until we arrive at the function of that&#x27;s given you. So the first graph is just here is just a rough sketch of a problem y equals x squared and that here were multiplying this by negative one. So this will take every Y value on our original graph and multiplied by negative one. So, for example, if X is one or minus one, the Y value was originally one. But when we multiply that by negative one, those y values will become minus one and the Vertex zeros the y values zero. So if you multiply that minus wanted still zero. So the second graf is just flips the problem upside down. And finally, the last transformation is to take our negative X square, and then we just add one to it. So if you want, you could think of this is negative X square, which was this over here and I were adding one to that. Why value So that&#x27;LL shift all over our why? Value&#x27;s up by one. So this negative one over here will get shifted up to zero. This y value over Hero zero will add one to that. It will now go upto one. So that&#x27;s over here. And this why I value when X was one the y values minus one. So we add one. It goes back to zero, and we could do a rough sketch here of our problem. And in this case, we got a bit lucky due to the function. As given, we don&#x27;t really have to do too much work to find the Vertex in the intercepts because these points already labeled on our graph we see that the X in ourselves, what she would normally have to find by just solving this equation here so you could write This is one minus X are actually in this case, Let&#x27;s given easier solution to this one. Add the X square to the other side, and then you see that X has to be plus or minus one, and those are points over here, so the intercepts will be negative. One comma, zero, one, comma zero and then the Vertex. We could also see from the graph. That&#x27;s just the point zero one. Usually we would have to do negative B over two A. That gives you the X value of the vortex and our problem. Over here, there&#x27;s no be turned because X to the one. The first power doesn&#x27;t appear. So be Zira. Plugging that in over here with a equals minus one gives you X equals zero and then plugging zero into F gives you one minus zero square equals one. So that&#x27;s why the Vertex is your old one. And we know that the access of symmetry is always given by the following equation. You said X equal to this quantity over here X value of the Vertex, and we just saw that that was a hero. So the equation for the line the access of symmetry is X equals zero so that it will just be this line over here. Don&#x27;t y axis. So it&#x27;s quite intro leading and green, and this is our X equals zero. This&#x27;s the access of symmetry. So we&#x27;ve used the sequence of transformations to graph the final answer. And we&#x27;ve identified the Vertex, the ex center sense and the access of symmetry, so that&#x27;s our final answer.

J Hardin · Answer

let&#x27;s go ahead and sketch this function many ways to do this I&#x27;Ll just go ahead and put this in the standard form so I&#x27;ll complete the square here so I&#x27;ll go ahead and take this negative one half in front of the ex and then we divide that by two and then you square the result However, notice that that&#x27;s being multiplied by two So we&#x27;ll make up for by subtracting two over sixteen Okay, so this gives us something information It tells us that the Vertex is the point one four comma seven eight So perhaps in the eggs direction we can go by one half and then in the UAE we&#x27;Ll just go buy one is normal So our vertex is at the point one fourth comma seven eight So that&#x27;s a brown right over here somewhere also accesses imagery You just take the X coordinate your vertex and write X equals that that value We could even plot the graft as well that she&#x27;s the rural coal line passing through one over four Okay, so we have the Vertex access the symmetry and I guess at this point we could just complete the rest of the graph, so notice that you could obtain our graph from the graph of X squared. You shifted to the right by one over four stretch of Brinkley by a factor of two and then shifted upward by seven over eight. So this craft will look very similar to a usual problem. But it&#x27;s Brinkley stretched by a factor of two. So at this point, you can just use this fact and give ah, rough looking sketch. If you want a more akra graph, you could plug in some more points like one half one and so on. And also we see that the Graf has no exit triceps. You could either see that visually from the grass or the other way to obtain it would be to take your quadratic, said it equal to zero and try to solve for X. But if you do, you see that there&#x27;s no solution because the left hand side positive number of times a positive number is bigger than your equal to zero, whereas the other side is strictly less than zero. So no solution

J Hardin · Answer

is that we can grab f I just starting with the graph of X squared and doing a few steps here, and it&#x27;s important to know which each step does photograph of the previous function. So here we take our original problem. Multiply by one over for so this is a vertical stretch by a factor of one over four. Here you see the negative sign. So we flipped the graph about the X axis and then finally the plus sixteen. This is sixteen units upward in the vertical direction. So on the side, let&#x27;s just do a little rough sketch of what&#x27;s going on. You have your X Square, multiply that by one over for that, makes it a flatter and then put the negative on the one over four that flips the graph about the X axis and then finally adding sixteen will just shift the latest graph upward by sixteen. So if we start off at the origin here when x zero, we see that we end up at sixteen and small rabbit, which is good and plugging a point or two. If you plug in one for X. Okay, you can see that you&#x27;re getting something that&#x27;s very close to sixteen now. Claudia, too. Then you&#x27;re getting sixteen minus one fifteen. So that&#x27;s a notable difference. And you can see here, by the way we&#x27;ve drawn the grids at this graph will look really flat, and we&#x27;LL come back and continue this after identifying the ex intercepts. But the Vertex we can already see it. The Vertex is just zero sixteen, which implies that the axis of symmetry is X equals zero. Now for the ex intercepts, as usual, just take your F set it equal to zero and softer X. So this will give you X squared people sixty four. And that means excess plus or minus eight. So this graph will not come down into the X axis until we&#x27;re at eight or minus eight, some more or less from sketch here. So here&#x27;s negative, eh? And plus eight. And there&#x27;s a rough sketch of our function. F

Problem

In Exercises 17-34, sketch the graph of the quadr…

Problem 33 Easy Difficulty

In Exercises 17-34, sketch the graph of the quadratic function without using a graphing utility. Identify the vertex, axis of symmetry, and x-intercept(s).

$ f(x) = \frac{1}{4} x^2 - 2x - 12 $

Answer

Vertex: $(4,-16)$
Axis of symmetry: $x=4$
$x$ -intercepts: $(12,0),(-4,0)$

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Catherine R.

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Vertex: $(4,-16)$Axis of symmetry: $x=4$$x$ -intercepts: $(12,0),(-4,0)$

Precalculus with Limits

Catherine R.

Anna Marie V.

Alayna H.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

Ron LarsonPrecalculus with Limits

Catherine R.

Anna Marie V.

Alayna H.

Caleb E.

Vertex: $(4,-16)$
Axis of symmetry: $x=4$
$x$ -intercepts: $(12,0),(-4,0)$

Ron Larson

Precalculus with Limits